Chapter 18.3, Problem 18.140P

To determine

(a)

The maximum value of the angle $\theta$ in the ensuing motion.

Answer to Problem 18.140P

The maximum value of the angle $\theta$ in the ensuing motion is $76.290°$.

Explanation of Solution

Given information:

The height of the solid cone is $180\text{mm}$, the radius of the circular cone is $60\text{mm}$, the position angle is $30°$, the rate of the spin is $300\text{rad}/\text{s}$ and the rate of the precession is $-4\text{rad}/\text{s}$.

Write the expression for the height of the center of gravity from the apex.

$C=\frac{3}{4}h$ ...... (I)

Here, the height of the solid cone is $h$.

Write the expression for the mass moment of the inertia.

$I=\frac{3}{10}m{r}^2$ ...... (II)

Here, the mass of the solid cone is $m$ and the radius of the solid cone is $r$.

Write the expression for the mass moment of the inertia in the transverse axis.

$I^{\prime }=\frac{3}{20}m{r}^2+\frac{3}{5}m{h}^2$ ...... (III)

Write the expression for the angular velocity in the x- direction.

${\omega }_x=-{\dot{\varphi }}_o\sin {\theta }_o$ ...... (IV)

Here, the rate of the precession is ${\dot{\varphi }}_o$ and the position angle is ${\theta }_o$.

Write the expression for the angular velocity in y- direction.

${\omega }_y={\dot{\theta }}_o$ ...... (V)

Here, the rate of the change of the angle is ${\dot{\theta }}_o$.

Write the expression for the angular velocity in z- direction.

${\omega }_z={\dot{\psi }}_o+{\dot{\varphi }}_o\sin {\theta }_o$ ...... (VI)

Here, the rate of the spin is ${\dot{\psi }}_o$.

Write the expression for the kinetic energy.

$\frac{T}{m}=\frac{1}{2}\frac{I^{\prime }}{m}\left({\omega }_x^2+{\omega }_y^2\right)+\frac{1}{2}\frac{I}{m}{\omega }_z^2$ ...... (VII)

Write the expression for the potential energy.

$\frac{V}{m}=gC\cos {\theta }_o$ ...... (VIII)

Write the expression for the total energy.

$\frac{E}{m}=\frac{T}{m}+\frac{V}{m}$ ...... (IX)

Write the expression for the constant $\beta$.

$\frac{\beta }{m}=\frac{I{\omega }_z}{m}$ ...... (X)

Write the expression for the constant $\alpha$.

$\frac{\alpha }{m}=\frac{I^{\prime }}{m}{\dot{\varphi }}_o{\sin }^2{\theta }_o+\frac{\beta }{m}\cos {\theta }_o$ ...... (XI)

Write the expression for the force.

$\begin{array}{l}F\left(x\right)=\left[\frac{2E}{m}-\frac{\frac{{\beta }^2}{m^2}}{\frac{I}{m}}-2gC\right]\left(1-x^2\right)-\frac{m}{I^{\prime }}{\left(\frac{\alpha }{m}-\frac{\beta x}{m}\right)}^2\\ 0=\left[\frac{2E}{m}-\frac{\frac{{\beta }^2}{m^2}}{\frac{I}{m}}-2gC\right]\left(1-x^2\right)-\frac{m}{I^{\prime }}{\left(\frac{\alpha }{m}-\frac{\beta x}{m}\right)}^2\end{array}$ ...... (XII)

Here, the distance along the x -axis is $x$.

Write the expression for the maximum angle.

${\theta }_{\max }={\cos }^{-1}\left(x_{\min }\right)$ ...... (XIII)

Calculation:

Substitute $180\text{mm}$ for $h$ in Equation (I).

$\begin{array}{c}C=\frac{3}{4}\left(180\text{mm}\right)\\ =\frac{3}{4}\left(180\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)\\ =0.135\text{m}\end{array}$

Substitute $60\text{mm}$ for $r$ in Equation (II).

$\begin{array}{l}I=\frac{3}{10}m{\left(60\text{mm}\right)}^2\\ \frac{I}{m}=\frac{3}{10}{\left(60\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2\\ \frac{I}{m}=1.08\times {10}^{-3}{\text{m}}^2\end{array}$

Substitute $180\text{mm}$ for $h$ and $60\text{mm}$ for $r$ in Equation (III).

$\begin{array}{l}{I}^{\prime }=\frac{3}{20}m{\left(60\text{mm}\right)}^2+\frac{3}{5}m{\left(180\text{mm}\right)}^2\\ \frac{I^{\prime }}{m}=\frac{3}{20}{\left(60\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2+\frac{3}{5}{\left(180\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2\\ \frac{I^{\prime }}{m}=5.4\times {10}^{-4}{\text{m}}^2+0.01944{\text{m}}^2\\ \frac{I^{\prime }}{m}=19.98\times {10}^{-3}{\text{m}}^2\end{array}$

Substitute $30°$ for ${\theta }_o$ and $-4\text{rad}/\text{s}$ for ${\dot{\varphi }}_o$ in Equation (IV).

$\begin{array}{c}{\omega }_x=-\left(-4\text{rad}/\text{s}\right)\sin \left(30°\right)\\ =\left(4\text{rad}/\text{s}\right)\left(0.5\right)\\ =2\text{rad}/\text{s}\end{array}$

Substitute $0$ for ${\dot{\theta }}_o$ in Equation (V).

${\omega }_y=0$

Substitute $300\text{rad}/\text{s}$ for ${\dot{\psi }}_o$, $-4\text{rad}/\text{s}$ for ${\dot{\varphi }}_o$ and $30°$ for ${\theta }_o$ in Equation (VI)

$\begin{array}{c}{\omega }_z=\left(300\text{rad}/\text{s}\right)+\left(-4\text{rad}/\text{s}\right)\cos 30°\\ =\left(300\text{rad}/\text{s}\right)+\left(-4\text{rad}/\text{s}\right)\left(0.8660\right)\\ =300\text{rad}/\text{s}-3.464\text{rad}/\text{s}\\ =296.53\text{rad}/\text{s}\end{array}$

Substitute $2\text{rad}/\text{s}$ for ${\omega }_x$, $0$ for ${\omega }_y$, $296.53\text{rad}/\text{s}$ for ${\omega }_z$, $19.98\times {10}^{-3}{\text{m}}^2$ for $\frac{I^{\prime }}{m}$ and $1.08\times {10}^{-3}{\text{m}}^2$ for $\frac{I}{m}$ in Equation (VII).

$\begin{array}{c}\frac{T}{m}=\frac{1}{2}\left(19.98\times {10}^{-3}{\text{m}}^2\right)\left({\left(2\text{rad}/\text{s}\right)}^2+{\left(0\right)}^2\right)+\frac{1}{2}\left(1.08\times {10}^{-3}{\text{m}}^2\right){\left(296.53\text{rad}/\text{s}\right)}^2\\ =\frac{1}{2}\left(19.98\times {10}^{-3}{\text{m}}^2\right)\left(4{\text{rad}}^2/{\text{s}}^2\right)+\frac{1}{2}\left(1.08\times {10}^{-3}{\text{m}}^2\right)\left(87933.53{\text{rad}}^2/{\text{s}}^2\right)\\ =3.996\times {10}^{-3}{\text{m}}^2/{\text{s}}^2+47484.106\times {10}^{-3}{\text{m}}^2/{\text{s}}^2\\ =47.451{\text{m}}^2/{\text{s}}^2\end{array}$

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.135\text{m}$ for $C$ and $30°$ for ${\theta }_o$ in Equation (VIII).

$\begin{array}{c}\frac{V}{m}=\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.135\text{m}\right)\cos 30°\\ =\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.135\text{m}\right)\left(0.8660\right)\\ =1.146{\text{m}}^2/{\text{s}}^{\text{2}}\end{array}$

Substitute $1.146{\text{m}}^2/{\text{s}}^{\text{2}}$ for $\frac{V}{m}$ and $47.451{\text{m}}^2/{\text{s}}^{\text{2}}$ for $\frac{T}{m}$ in Equation (IX).

$\begin{array}{c}\frac{E}{m}=47.451{\text{m}}^2/{\text{s}}^{\text{2}}+1.146{\text{m}}^2/{\text{s}}^{\text{2}}\\ =48.597{\text{m}}^2/{\text{s}}^{\text{2}}\end{array}$

Substitute $296.53\text{rad}/\text{s}$ for ${\omega }_z$ and $1.08\times {10}^{-3}{\text{m}}^2$ for $\frac{I}{m}$ in Equation (X).

$\begin{array}{c}\frac{\beta }{m}=\left(296.53\text{rad}/\text{s}\right)\left(1.08\times {10}^{-3}{\text{m}}^2\right)\\ =0.3202{\text{m}}^{\text{2}}/\text{s}\end{array}$

Substitute $19.98\times {10}^{-3}{\text{m}}^2$ for $\frac{I^{\prime }}{m}$, $0.3202{\text{m}}^{\text{2}}/\text{s}$ for $\frac{\beta }{m}$, $-4\text{rad}/\text{s}$ for ${\dot{\varphi }}_o$ and $30°$ for ${\theta }_o$ in Equation (XI).

$\begin{array}{c}\frac{\alpha }{m}=\left(19.98\times {10}^{-3}{\text{m}}^2\right)\left(-4\text{rad}/\text{s}\right){\sin }^2\left(30°\right)+\left(0.3202{\text{m}}^{\text{2}}/\text{s}\right)\cos \left(30°\right)\\ =\left(19.98\times {10}^{-3}{\text{m}}^2\right)\left(-4\text{rad}/\text{s}\right)\left(0.25\right)+\left(0.3202{\text{m}}^{\text{2}}/\text{s}\right)\left(0.8660\right)\\ =-0.01998{\text{m}}^{\text{2}}/\text{s}+0.2772{\text{m}}^{\text{2}}/\text{s}\\ =0.2572{\text{m}}^{\text{2}}/\text{s}\end{array}$

Substitute $1.08\times {10}^{-3}{\text{m}}^2$ for $\frac{I}{m}$, $0.2572{\text{m}}^{\text{2}}/\text{s}$ for $\frac{\alpha }{m}$, $0.3202{\text{m}}^{\text{2}}/\text{s}$ for $\frac{\beta }{m}$, $19.98\times {10}^{-3}{\text{m}}^2$ for $\frac{I^{\prime }}{m}$, $48.597{\text{m}}^2/{\text{s}}^{\text{2}}$ for $\frac{E}{m}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.135\text{m}$ for $C$ in Equation (XII).

$\begin{array}{l}0=\left\{\begin{array}{l}\left[\begin{array}{l}2\left(48.597{\text{m}}^2/{\text{s}}^{\text{2}}\right)-\frac{{\left(0.3202{\text{m}}^{\text{2}}/\text{s}\right)}^2}{\left(1.08\times {10}^{-3}{\text{m}}^2\right)}-\\ 2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.135\text{m}\right)\end{array}\right]\left(1-x^2\right)-\\ \left(\frac{1}{19.98\times {10}^{-3}{\text{m}}^2}\right){\left(\left(0.2572{\text{m}}^{\text{2}}/\text{s}\right)-\left(0.3202{\text{m}}^{\text{2}}/\text{s}\right)x\right)}^2\end{array}\right\}\\ \left\{\begin{array}{l}\left[\left(97.194{\text{m}}^2/{\text{s}}^{\text{2}}\right)-94.93{\text{m}}^2/{\text{s}}^{\text{2}}-2.648{\text{m}}^2/{\text{s}}^{\text{2}}\right]\left(1-x^2\right)-\\ \left(0.1986{\text{m}}^2/{\text{s}}^{\text{2}}{x}^2\right)\end{array}\right\}=0\\ \left\{\left(0.384{\text{m}}^2/{\text{s}}^{\text{2}}\right)\left(1-x^2\right)-\left(\left(0.1986{\text{m}}^2/{\text{s}}^{\text{2}}\right)x^2\right)\right\}=0\\ x=0.237,0.860,1.73\end{array}$

Select the minimum value of $x$ for the maximum angle.

Substitute $0.237$ for $x_{\min }$ in Equation (XIII).

$\begin{array}{c}{\theta }_{\max }={\cos }^{-1}\left(0.237\right)\\ =76.290°\end{array}$

Conclusion:

The maximum value of the angle $\theta$ in the ensuing motion is $76.290°$.

To determine

(b)

The corresponding value of the rate of precession.

The corresponding value of the rate of spin.

Answer to Problem 18.140P

The corresponding value of the rate of precession is $9.341\text{rad}/\text{s}$.

The corresponding value of the rate of spin is $294.316\text{rad}/\text{s}$.

Explanation of Solution

Write the expression for the rate of the precession.

$\dot{\varphi }=\frac{\frac{\alpha }{m}-\frac{\beta }{m}\cos \theta }{\frac{I^{\prime }}{m}{\sin }^2\theta }$ ...... (XIV)

Write the expression for the rate of the spin.

$\dot{\psi }={\omega }_z-\dot{\varphi }\cos \theta$ ...... (XV)

Calculation:

Substitute $0.2572{\text{m}}^{\text{2}}/\text{s}$ for $\frac{\alpha }{m}$, $0.3202{\text{m}}^{\text{2}}/\text{s}$ for $\frac{\beta }{m}$, $19.98\times {10}^{-3}{\text{m}}^2$ for $\frac{I^{\prime }}{m}$ and $76.290°$ for $\theta$ in Equation (XIV).

$\begin{array}{c}\dot{\varphi }=\frac{\left(0.2572{\text{m}}^{\text{2}}/\text{s}\right)-\left(0.3202{\text{m}}^{\text{2}}/\text{s}\right)\cos \left(76.290°\right)}{\left(19.98\times {10}^{-3}{\text{m}}^2\right){\sin }^2\left(76.290°\right)}\\ =\frac{\left(0.2572{\text{m}}^{\text{2}}/\text{s}\right)-\left(0.3202{\text{m}}^{\text{2}}/\text{s}\right)\left(0.2370\right)}{\left(19.98\times {10}^{-3}{\text{m}}^2\right)\left(0.9715\right)}\\ =\frac{\left(0.1813{\text{m}}^{\text{2}}/\text{s}\right)}{\left(19.41\times {10}^{-3}{\text{m}}^2\right)}\\ =9.341\text{rad}/\text{s}\end{array}$

Substitute $9.341\text{rad}/\text{s}$ for $\dot{\varphi }$, $296.53\text{rad}/\text{s}$ for ${\omega }_z$ and $76.290°$ for $\theta$ in Equation (XV).

$\begin{array}{c}\dot{\psi }=296.53\text{rad}/\text{s}-\left(9.341\text{rad}/\text{s}\right)\cos \left(76.290\right)\\ =296.53\text{rad}/\text{s}-\left(2.213\text{rad}/\text{s}\right)\\ =294.316\text{rad}/\text{s}\end{array}$

Conclusion:

The corresponding value of the rate of precession is $9.341\text{rad}/\text{s}$.

The corresponding value of the rate of spin is $294.316\text{rad}/\text{s}$.

To determine

(c)

The value of the angle at which the sense of precession is reversed.

Answer to Problem 18.140P

The value of the angle at which the sense of precession is reversed is $36.55°$.

Explanation of Solution

Write the expression for the rate of the precession.

$\begin{array}{l}\dot{\varphi }=\frac{\frac{\alpha }{m}-\frac{\beta }{m}\cos \theta }{\sin \theta }\\ \frac{\frac{\alpha }{m}-\frac{\beta }{m}\cos \theta }{\sin \theta }=0\\ \frac{\alpha }{m}-\frac{\beta }{m}\cos \theta =0\\ \cos \theta =\frac{\frac{\alpha }{m}}{\frac{\beta }{m}}\end{array}$ ...... (XVI)

Calculation:

Substitute $0.2572{\text{m}}^{\text{2}}/\text{s}$ for $\frac{\alpha }{m}$ and $0.3202{\text{m}}^{\text{2}}/\text{s}$ for $\frac{\beta }{m}$ in Equation (XVI).

$\begin{array}{l}\cos \theta =\frac{0.2572{\text{m}}^{\text{2}}/\text{s}}{0.3202{\text{m}}^{\text{2}}/\text{s}}\\ \theta ={\cos }^{-1}\left(0.8032\right)\\ \theta =36.55°\end{array}$

Conclusion:

The value of the angle at which the sense of precession is reversed is $36.55°$.